A testing laboratory wants to determine if a new widget can withstand large acceleration and decelerations. To find out, they glue a 5.0kg widget to a test stand that will drive it vertically up and down. At 0 seconds the widget is at an acceleration of 19.6m/s^2. It reaches 0 at 1/2 a second and drops to -19.6m/s^2 by 1 sec The widget decelerates at 39.2m/s^2 every second.
a) Identify the forces acting on the widget and draw a free body diagram.
I'm already stuck here. I know there's a normal force and a gravitational force.
b) Determine the value of ny, the y-component of the normal force acting on the widget during the first second of motion. Give your answer as a graph of ny versus t.
Why is there a y-component to the normal force? Since it's accelerating vertically wouldn't there be no x-component, therefore the y-component would be the whole normal force?
a) The forces acting on the widget are the gravitational force (weight) and the normal force.
The gravitational force is always acting downward and is given by the equation Fg = mg, where m is the mass of the widget and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force acts perpendicular to the surface of contact between the widget and the test stand. In this case, since the widget is being driven vertically up and down, the normal force will vary with time. In the up direction, the normal force will be less than the weight of the widget, and in the down direction, the normal force will be greater than the weight of the widget.
b) The y-component of the normal force refers to the vertical component of the normal force. Since the widget is being driven vertically up and down, there is no x-component of the normal force. However, the y-component changes with time due to the acceleration of the widget.
To determine the value of ny, we can analyze the motion of the widget during the first second using the given information:
- At 0 seconds, the widget has an acceleration of 19.6 m/s^2 upward. This means the normal force is less than the weight of the widget.
- At 0.5 seconds, the widget reaches a velocity of 0. This means the net force acting on the widget is equal to 0. The normal force at this point is equal to the weight of the widget.
- At 1 second, the widget has a deceleration of 39.2 m/s^2 downward. This means the normal force is greater than the weight of the widget.
To graph ny versus t during the first second, we can use the following information:
- From t = 0 to 0.5 seconds, ny is less than the weight of the widget.
- At t = 0.5 seconds, ny is equal to the weight of the widget.
- From t = 0.5 to 1 second, ny is greater than the weight of the widget.
Please note that without specific numeric values (e.g., actual weight of the widget), we cannot provide an exact graph of ny versus t. However, the general trend would be a decrease in ny from t = 0 to 0.5 seconds, followed by a constant ny at t = 0.5 seconds, and then an increase in ny from t = 0.5 to 1 second.
a) You are correct that there are two main forces acting on the widget during the motion: the gravitational force and the normal force. The gravitational force, denoted by mg (where m is the mass of the widget and g is the acceleration due to gravity), always acts vertically downward. The normal force, denoted by N, acts perpendicular to the surface of contact (in this case, the test stand).
In a free body diagram, we represent these forces as vectors. Since the widget is being driven vertically up and down, the normal force will have both x and y components. The y-component of the normal force opposes the gravitational force, while the x-component is usually negligible in this scenario.
b) The y-component of the normal force, denoted by ny, can be determined by subtracting the gravitational force from the total normal force (N). Since the widget goes through various accelerations and decelerations, the value of ny may change over time.
To determine the value of ny throughout the first second of motion, we need to consider the acceleration of the widget and apply Newton's second law: ΣF = m⋅a.
At t = 0 seconds, the acceleration is 19.6 m/s^2 upwards. Therefore, the net force acting on the widget must be the sum of the gravitational force and the y-component of the normal force:
ΣF = mg - ny = m⋅a
We can rearrange this equation to solve for ny:
ny = mg - m⋅a
Since the acceleration is constant during this first second, we can substitute a = 19.6 m/s^2 to get:
ny = m⋅(g - a)
The graph of ny versus t can now be plotted using the equation above. It will be a straight line, starting at mg (the full gravitational force) and decreasing at a constant rate of m⋅a.